Question: $\dfrac{ -5e + 7f }{ 5 } = \dfrac{ -6e + g }{ 7 }$ Solve for $e$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ -5e + 7f }{ {5} } = \dfrac{ -6e + g }{ 7 }$ ${5} \cdot \dfrac{ -5e + 7f }{ {5} } = {5} \cdot \dfrac{ -6e + g }{ 7 }$ $-5e + 7f = {5} \cdot \dfrac { -6e + g }{ 7 }$ Multiply both sides by the right denominator. $-5e + 7f = 5 \cdot \dfrac{ -6e + g }{ {7} }$ ${7} \cdot \left( -5e + 7f \right) = {7} \cdot 5 \cdot \dfrac{ -6e + g }{ {7} }$ ${7} \cdot \left( -5e + 7f \right) = 5 \cdot \left( -6e + g \right)$ Distribute both sides ${7} \cdot \left( -5e + 7f \right) = {5} \cdot \left( -6e + g \right)$ $-{35}e + {49}f = -{30}e + {5}g$ Combine $e$ terms on the left. $-{35e} + 49f = -{30e} + 5g$ $-{5e} + 49f = 5g$ Move the $f$ term to the right. $-5e + {49f} = 5g$ $-5e = 5g - {49f}$ Isolate $e$ by dividing both sides by its coefficient. $-{5}e = 5g - 49f$ $e = \dfrac{ 5g - 49f }{ -{5} }$ Swap signs so the denominator isn't negative. $e = \dfrac{ -{5}g + {49}f }{ {5} }$